Question: Let $a,$ $b,$ $c$ be real numbers such that $a + b + c = 0.$  Find the set of all possible values of $ab + ac + bc.$
Answer: Squaring the equation $a + b + c = 0,$ we get
\[a^2 + b^2 + c^2 + 2(ab + ac + bc) = 0.\]Hence, $2(ab + ac + bc) = -(a^2 + b^2 + c^2) \le 0,$ so
\[ab + ac + bc \le 0.\]Equality occurs when $a = b = c = 0.$

Now, set $c = 0,$ so $a + b = 0,$ or $b = -a.$  Then
\[ab + ac + bc = ab = -a^2\]can take on all nonpositive values.  Therefore, the set of all possible values of $ab + ac + bc$ is $\boxed{(-\infty,0]}.$